To solve for the smallest number of terms \( n \) required such that the sum \( S_n \) of the arithmetic series exceeds 10000, we can use the formula for the sum of the first \( n \) terms of an arithmetic series:
\[ S_n = \frac{n}{2} \times [2a_1 + (n - 1)d] \]
where:
- \( S_n \) is the sum of the first \( n \) terms,
- \( a_1 \) is the first term of the series,
- \( d \) is the common difference between the terms,
- \( n \) is the number of terms.
From the series provided, we can identify the first term \( a_1 = 325 \) and the common difference \( d = 350 - 325 = 25 \).
We want to find the smallest \( n \) such that \( S_n > 10000 \). Replacing \( a_1 \) and \( d \) in the sum formula we get:
\[ S_n = \frac{n}{2} \times [2 \times 325 + (n - 1) \times 25] \]
Now we can expand and simplify the equation:
\[ S_n = \frac{n}{2} \times [650 + 25n - 25] \]
\[ S_n = \frac{n}{2} \times [625 + 25n] \]
\[ S_n = \frac{n}{2} \times 25 \times [25 + n] \]
\[ S_n = 12.5n \times [25 + n] \]
Now we need \( S_n > 10000 \), which gives us:
\[ 12.5n \times [25 + n] > 10000 \]
Dividing both sides by 12.5 to simplify:
\[ n \times [25 + n] > 800 \]
We now have a quadratic inequality in the form:
\[ n^2 + 25n - 800 > 0 \]
We have to find the value of \( n \) that satisfies this inequality. We will find the roots of the corresponding quadratic equation:
\[ n^2 + 25n - 800 = 0 \]
To find the roots, we can use the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the quadratic equation \( an^2 + bn + c = 0 \), the coefficients are \( a = 1 \), \( b = 25 \), and \( c = -800 \). Plugging these values into the quadratic formula:
\[ n = \frac{-25 \pm \sqrt{25^2 - 4 \times 1 \times (-800)}}{2 \times 1} \]
\[ n = \frac{-25 \pm \sqrt{625 + 3200}}{2} \]
\[ n = \frac{-25 \pm \sqrt{3825}}{2} \]
\[ n = \frac{-25 \pm 61.9}{2} \]
We get two solutions for \( n \):
1. \( n = \frac{-25 - 61.9}{2} \) which gives a negative value, which is not acceptable as we cannot have a negative number of terms.
2. \( n = \frac{-25 + 61.9}{2} \) which gives a positive value.
Calculating the positive root:
\[ n = \frac{-25 + 61.9}{2} \]
\[ n = \frac{36.9}{2} \]
\[ n = 18.45 \]
Since \( n \) has to be an integer because we cannot have a fraction of a term, and we're looking for the smallest \( n \) that meets the condition, we round up the result (as the sum must exceed 10000).
Thus, the smallest integer greater than 18.45 is 19.
Therefore, the smallest number of terms required for the sum of the arithmetic series to exceed 10000 is \( n = 19 \).
